Timeline for What is the direct coupling problem which differential amplifiers help avert?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 29 at 21:35 | comment | added | EE18 | Awesome, thank you Andy! | |
| Jan 29 at 21:31 | comment | added | Andy aka | @EE18 see this also also. It's all about balancing impedances so that noise affects both wires the same. | |
| Jan 29 at 21:09 | comment | added | EE18 | As you further point out, in a single-ended directly-coupled arrangement, even if we could arrange for perfectly equal output bias voltage of one stage with input bias voltage of the next, slight noise sources would be propagated and cause potential hitting of end stops. In contrast, I guess our hope with differential circuits is that even if noise arises, it arises equally on each input so that no noise is propagated. This second property is referenced in the opening paragraph to Chapter 3.5 of the text (for posterity and my future reference!) along with the biasing problem alluded to above. | |
| Jan 29 at 21:07 | comment | added | EE18 | ...we will have some direct current current flowing between the stages at worst or, at best, the given stage of interest will see this delta in biasing at the input as a signal input which will, as you say, cause that stage to "hit the end-stops." The crucial property of the differential amplifier of sufficient symmetry is that if it is biased with \$V_{id} = 0\$ then its output will have \$V_{od} = 0\$. If this output is in turn used as the bias for a subsequent differential stage then we have \$V_{id} = 0\$ for that stage, as is ideal. | |
| Jan 29 at 21:04 | comment | added | EE18 | Just returning here in light of this question (electronics.stackexchange.com/questions/698547/…) as well as a reread of the given section from Gray and Hurst. I think in retrospect I somewhat missed your initial point which is crucial, as given by you in your line "You could not connect this directly to another similar stage because that extra stage wants to see 0 volts as a neutral input...." I guess the initial point you're making here is that, with direct coupling (as is the case here), if a previous stage's biasing does not match up with our stage of interest's biasing then, in effect... | |
| Nov 28, 2023 at 17:12 | vote | accept | EE18 | ||
| Nov 28, 2023 at 14:37 | history | edited | Andy aka | CC BY-SA 4.0 |
added 1067 characters in body
|
| Nov 28, 2023 at 14:27 | comment | added | Andy aka | Correct @EE18. I'll try and show an example regarding your first comment.... | |
| Nov 28, 2023 at 14:14 | comment | added | EE18 | ...While in contrast, going differentially means that (if we have little common mode gain) then we'll only have signal gain and thus won't have this issue? | |
| Nov 28, 2023 at 14:14 | comment | added | EE18 | I am still not sure I see, and perhaps that's because I don't follow what would go wrong if we simply cascaded non-differential (is this the term?) amplifiers. What goes wrong if I directly couple non-differential amplifiers? Are you saying that spurious sources (which I think are what generate common mode gain?) would add on the output to signal gain and then put the input to the next amplifier out of its linear range of operation (is this what you mean by saturate?)?... | |
| Nov 28, 2023 at 13:40 | history | answered | Andy aka | CC BY-SA 4.0 |